20 Answers
20

One way to find the coefficients, assuming we already know that it's a degree $3$ polynomial, is to calculate the sum for $n=0,1,2,3$. This gives us four values of a degree $3$ polynomial, and so we can find it.

The better way to approach it, though, is through the identity
$$ \sum_{t=0}^n \binom{t}{k} = \binom{n+1}{k+1}. $$
This identity is true since in order to choose a $(k+1)$-subset of $n+1$, you first choose an element $t+1$, and then a $k$-subset of $t$.

That's a good method: +1! Concerning the polynomial approach, I think it could be enhanced by noticing that if $p_k(n)=\sum_{i=1}^n i^k$ then, extending what you wrote, $p_k\in\mathbb Q[n]$, $\partial p_k=k+1$, and $n(n+1)\,\Big|\,p_k$ for every $k$. For $k=2$ this reduces to 2 the coefficients to be found making this approach equally worth, isn't it?
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AndreasTMar 30 '13 at 15:58

To discover the identity, notice that any polynomial solution of the above recurrence has degree at most $3$. Hence it's easy to find the polynomial solution by substituting a cubic polynomial with undetermined coefficients.

Sums of polynomials can be done completely mechanically (no insight required, just turn the handle!) using the discrete calculus. Bill Dubuque mentions this in his answer, but I think it's nice to see a worked example.

A natural approach for this kind of problems when you don't know the result is to proceed as follows :

We may want to write the sum $\sum_{k=1}^n k^2$ as a telescopic sum, so we will try to find a polynomial of degree 3 ( why ? ) $P$ so that $P\left( k+1 \right) - P\left(k\right)=k^2$. Let $P\left( x \right) = ax^3+bx^2+cx$ for all reals $x$, then our constraint becomes :

Similarly, we can derive for each $k$
$$ \sum_{j=0}^{k-1} {k \choose j} E(X^j) = \sum_{l=1}^k {k \choose l} n^{l-1} $$
and so if we know $E(X^0), \ldots, E(X^{k-2})$ we can solve for $E(X^{k-1})$. So this method generalizes to higher moments as well.

I think it's useful to report here another proof that I have posted on Mathoverflow.

Write down numbers in an equilateral triangle as follows:

1
2 2
3 3 3
4 4 4 4

Now, clearly the sum of the numbers in the triangle is $Q_n:=1^2+2^2+\dots+n^2$. On the other hand, if you superimpose three such triangles rotated by $120^\circ$ each, like these ones

1 4 4
2 2 3 4 4 3
3 3 3 2 3 4 4 3 3
4 4 4 4 1 2 3 4 4 3 2 1

then the sum of the numbers in each position equals $2n+1$. Therefore, you can double-count $3Q_n=\frac{n(n+1)}{2}(2n+1)$. $\square$

The proof is not mine and I do not claim otherwise. I first heard it from János Pataki. It is similar (but simpler) to the proof appearing on Wikipedia as I am writing this.

How to prove formally that all positions sum to $2n+1$? Easy induction: moving down-left or down-right from the topmost number does not alter the sum, since one of the three summand increases and one decreases. This is a discrete analogue of the Euclidean geometry theorem "given a point $P$ in an equilateral triangle $ABC$, the sum of its three distances from the sides is constant" (proof: sum the areas of $APB,BPC,CPA$), which you can mention as well.

Let $S=\{1,2,\dots,(n+1)\},n\ge 2$ and $T=\{(x,y,z)|x.y,z\in S,x< z,y< z\}$.By counting the number of members of $T$ in $2$ different ways I will prove the formula.

$1$st way:

We will at first Choose $z$ form the set $S$.When $z$ is $1$ then there are no choices for $x,y$ so the no. of elements of $T$ with $z=0$ is zero.When $z=2$ the number of choices for $x$ is $1$ and so is for $y$(precisely $x=y=1$).When $z=3$ then $x\in \{1,2\}$ and $y\in \{1,2\}$ so total no. of choices equals $2^2$.In a similar manner when $z=k,(1\le k\le (n+1))$,no. of choices for $x$ equals $(k-1)$ and no. of choices for $y$ is also $(k-1)$. So total no . of elements of T with $z=k$ is $(k-1)^2$.

So we will get the total no. of elements of $T$ by summing $(k-1)^2$ up from $1 $ to $(n+1)$.Hence $$|T|=\sum _{l=1}^{(n+1)}(l-1)^2=\sum_{k=1}^{n}k^2$$

$2$nd way:

Among the elements of $T$ consisting of three numbers from the set $S$, there are elements in $x=y$ and elements in which $x\ne y$.

We can count the no. of elements in which $x=y$ by choosing two distinct nos. from $S$ and assigning $z$ with the lagest no. and $x,y$ with the smallest number. We can choose two distinct numbers from $S$ in $\displaystyle \binom{n+1}{2}$ ways, so the total no. elements having $x=y$ is $\displaystyle \binom{n+1}{2}$.

Now we have to count the number of elements in which $x\ne y$.This means that $x,y$ are dinstict and as they are less than $z$ this means that all the three are distinct. So we can count no. of such elements in $T$ in the following way.At first we will choose three elements from the set $S$ and assign the largest value to $z$ and assign the other two values to $x,y$. Now we can choose three numbers from the set $S$ in $\displaystyle \binom{n+1}{3}$.From each such three element we can get two elements of the set $T$(Assigning the largest to $z$ and then assigning any one of then to $x$ and the other to $y$). So no. of elements of $T$ having $x\ne y$ is $2\displaystyle \binom{n+1}{3}$

So by this method we have $|T|=\displaystyle \binom{n+1}{2}+2\displaystyle \binom{n+1}{3}$

On equating the result obtained from both the methods we have
$$\sum_{k=1}^{n}k^2=\displaystyle \binom{n+1}{2}+2\displaystyle \binom{n+1}{3}=\frac{n(n+1)(2n+1)}{6}$$

Note that this can easily be extended to find the sum of $p$th power of integers.($p\in \mathbb{N})$

Base case: If $n=0$, then we have $0$ on the left hand side, and $0(0+1)(2(0)+1)/6=0$ on the right.

Induction step:

Consider the differences $L(j+1)-L(j)$, and $R(j+1)-R(j)$ where $L(j)$ indicates that we have $j$ for $n$ on the left hand side. Well, $L(j+1)-L(j)=(j+1)^2$, and
$$R(j+1)-R(j)=\frac{(j+1)((j+1)+1))(2(j+1)+1)}{6} - \frac{j(j+1)(2j+1)}{6}$$ which simplifies to $(j+1)^2$ also. So, the rates of change on both sides equal each other, and thus the induction step follows.

The only "Gauss summation formula" a google search turns up is the formula $$\sum_{k=1}^nk=\frac{n(n+1)}{2}$$ which is not relevant to the OP's question and does not do what you said. The sum of the first $n$ $k$th powers is a complicated expression involving Bernoulli numbers. Also, I'm not sure what "guessing" you are referring to. Please improve your answer.
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Zev ChonolesJun 28 '11 at 5:19

@Zev Chonoles: So you think everything of value has been loaded onto the web? You have the faith of a Breton peasant. Anyway, the Bernoulli numbers are implicit in the formula that I cited, and I added a link giving the justification of the formula.
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Mike JonesJul 12 '11 at 20:19

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Where did I ever say I thought "everything of value has been loaded onto the web"? Since you had explained nothing about the mathematics of the claim in your answer, there was nothing to go on other than the words "Gauss's summation formula". It's not the internet's fault you were using a non-standard name for this formula. And I don't know what the faith of a Breton peasant is supposed to be like (this quote makes no sense to me), but I don't think I appreciate your implication, whatever it is.
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Zev ChonolesJul 12 '11 at 20:51

At any rate: yes, now that we have determined what formula it is you are referring to, I agree that the Bernoulli numbers are implicit in it.
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Zev ChonolesJul 12 '11 at 20:57

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Obviously, I have no control over the voting decisions of others, so I have no clue what it is you're complaining to me about. Furthermore, in my opinion people should choose how to vote on a post independently of what its current total score is; there is no such thing as an "appropriate" total score for a post. If three people came across your post and independently decided that it was not sufficiently useful, then a total score of -3 is the result.
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Zev ChonolesJul 14 '11 at 5:03